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Hcl Naoh Essay

Words: 527, Paragraphs: 1, Pages: 2

Paper type: Essay

Purpose of this lab was to prepare solutions of acid base and then standardize them by using the titration method. Then use them to determine the unknown concentration of an acid solution. Titration is a procedure that is used to study the stoichiometry of a reaction. Before we start the lab we have to some pre-lab exercises.The first exercise is to calculate the number of grams of NaOH required to make 250 mL of a .1 M solution of NaOH. When we calculate this we get 1g. Then the second exercise was to calculate the number of milliliters of 1.20 M HCL required to make 100.0 mL of a .100 M solutions of HCL. The answer was 8.33 ml.DATA-Standardization of NaOH SolutionSample 1Sample 2Sample 3Mass of KHP.335 g.320 g.307 gInitial Volume NaOH, Vi(mL)0 mL17.2 mL27.3 mLFinal Volume NaOH, Vf (mL)17.2 mL34.6 mL43.4 mLTotal Volume NaOH used (mL)17.2 mL17.4 ml16.1 mLMoles KHP.0016 mL.0016 mL.0015 mLMoles NaOH.0016 mol.0016 mol.0015 molMolarity NaOH.093 M.092 M.093 MAverage Molarity.093 M–Standardization of HCL SolutionSample 1Sample 2Sample 3Volume of HCL12 mL12 mL12 mLInitial Volume NaOH, Vi0 mL11.8 mL22.8 mLFinal Volume NaOH, Vf11.8 mL22.8 mL34.6 mLMolarity NaOH.093 M.093 M.093 mLMoles NaOH.0001 mol.0102 mol.00109 molMoles HCL.0019.0102 mol.00109 molMolarity of HCL.093 M.093 M.092 MAverage Morality.093 M–Concentration of Unknown HCL Solution……….Unknown 2Sample 1Sample 2Sample 3Volume of HCL12 mL12 mL12 mLInitial Volume NaOH, Vi (mL)Final Volume NaOH used (mL)16 mL21.2 mL26.6 mLTotal Volume NaOH, used (mL)15 mL5.2 mL5.4 mLMolarity NaOH.093 M.093 M.093 MMoles NaOH.000484 mol.000484 mol.000502 molMolarity of HCL.000484 mol.000484 mol.00502 molMolarity of HCL.040 M.040 M.042AVERAGE MOLARITY .041 MConclusion- The titration methods were use to prepare solutions of acid and base and also to determine the unknown concentration of an acid solution. Average Molarity of the unknown solution was .041 M.Supplementary Questions-1. Using the exact mass of NaOH measured and recorded at the beginning of the experiment, calculate the molarity 250 mL NaOH solution. How does this compare to the NaOH concentration you calculated using the data in part 4? What could be some reasons for this?1g * 1mol/40g * 1L/.25 L= .1 M It is greater then the concentration in part 4.2. Why were the amounts of distilled water added to the acids not precisely recorded and considered to be irrelevant? This might be because distilled water has no effect on the amount of the acid because the acid is stronger.3. Would the calculated molarity of the HCL solutions be higher or lower or not affected if each of the following occurred. Explain your answersA. The buret containing NaOH was rinsed with distilled water but not rinsed with NaOH before being filled.Reduce NaOH concentration, more NaOH would be used. Therefore molarity of HCL would be more.B. The calculated molarity was too highThe moles of NaOH used would be high. Therefore HCL would be highC. The tip of the NaOH buret contained an air bubble at the beginning of the titration but not at the end.It would seem like more NaOH was used Therefore HCL higher.4. Suppose you used the base Ca(OH)2 instead of NaOH. What changes would need be made to the calculations of the HCL concentrations?It would have to double because twice as more OH ions are used.

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